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marta [7]
2 years ago
15

a 10.0 kg sphere is released from rest in an ocean. as it falls, the water applies a resistive force r

Physics
1 answer:
dimaraw [331]2 years ago
4 0

The calculated coefficient of kinetic friction is 0.33125.'

The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.

given mass of the block=10 kg

spring constant k= 2250 Nm

now according to principal of conservation of energy we observe,

the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.

mgh= μ (mgl) +1/2 kx²

10 x 10 x 3= μ(600) +(1125) (0.09)

μ(600) =300 - 101.25

μ = 198.75÷600

μ =0.33125

The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)

Learn more about kinetic friction here-

brainly.com/question/13754413

#SPJ4

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A wooden rod of negligible mass and length 84.0 cm is pivoted about a horizontal axis through its center. A white rat with mass
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Answer:

v = 2.029 m/s

Explanation:

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Due to gravity, the bar oscillates and becomes vertical. The mass that occupies the lower position is the one with the highest torque. The one that reduces the potential energy (the system tends to the position of minimum energy). This is achieved if the mass that goes down is 0.6kg (that goes down 42cm) and the one that goes up is 0.2kg (goes up 42cm).

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Ei = Ki + Ui = 0.5*(m₁+m₂)*(0)² + (m₁+m₂)*9.8*(0) = 0 J

Ef = Kf + Uf

⇒ Kf = 0.5*(m₁+m₂)*v² = 0.5*(0.6+0.2)*v² = 0.4*v²

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⇒ Ef = Kf + Uf = 0.4*v² - 1.6464

Since

0 = 0.4*v² - 1.6464  ⇒  v = 2.029 m/s

v is the same value due to the wooden rod is pivoted about a horizontal axis through its center and the masses are on opposite ends.

v₁ = v₂ = v    ⇒  ω₁*R₁ = ω₂*R₂  ⇒ ω₁*R = ω₂*R  ⇒ ω₁ = ω₂ = ω

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